CH17 Invariants has_two_sum.py wrong solution/test? Specific problem **buggy test**

[p2327]

Hello

I am moving through the book and encountered what I believe is an oversight.

Buggy test The task: given a list of numbers A, find if two numbers exist in the list that add up to a value t.

The book implementation:

def has_two_sum(A: List[int], t: int) -> bool:
    i, j = 0, len(A) - 1

    while i <= j:
        if A[i] + A[j] == t:
            return True
        elif A[i] + A[j] < t:
            i += 1
        else:  # A[i] + A[j] > t.
            j -= 1
    return False

My implementation:

def has_two_sum(A: List[int], t: int) -> bool:
    seen = dict()

    for index, value in enumerate(A):
        complement = t - value
        if complement in seen:
            return True
        else:
            seen[value] = index
    return False

The test case I am having trouble with:

Test FAILED (  53/1005) [  28 us]
Arguments
        A:        [-2, -1, 1, 4]
        t:        8
Failure info
        expected: True
        result:   False

What happens with the book solution is that 4 is checked twice as A[i] and A[j] are checked for i = j = index = 3, giving a total of 8 and a True result. However if we go by what the task is asking this should not be the case: we can easily see that in the given array not two numbers sum up to 8.

I found that by changing the while clause to i < j this is corrected.

Not sure this is a mistake in the solution or I have not correctly understood what is asked?